How to solve ODE Solve the DE:
$$2y^2y''+2y(y')^2=1$$
Is it possible to solve this by implicit substitution i.e. let $v = y'$ and thus 
$$\frac{dv}{dy}v = \frac{1-2yv^2}{2y^2}$$
by the chain rule. And then from here let $z = \frac{v}{y}$? What if $y=0$?
 A: It has been observed in the comments that the equation can be rewritten in terms of the derivatives of $y^2$, namely as
$$
y(y^2)''=1
$$
Multiplication with $4y'$ allows one integration to
$$
((y^2)')^2=4y+C
$$
So it remains to integrate (using a different constant $C$)
$$
\frac{yy'}{\sqrt{C\pm y}}=1.
$$
A: Note that $y(x) \sim x^\alpha$ satisfies the ODE. After substitution, a 2nd order equation is found for $\alpha$. This gives you the two solutions.
A: Starting with $$2y^2y''+2y(y')^2=1$$ let us use the reverse relations between derivatives, that is to say, $$y'_x=\frac 1  {x'_y}\qquad , \qquad y''_x=-\frac{x''_y}{{x'_y}^3}$$ and replace in the original equation. This leads to $$\frac{2 y}{x'^2}-\frac{2 y^2 x''}{x'^3}=1$$ and the function $x$ disappeared.
So, now, define $x'=t$ which gives $$\frac{2 y}{t^2}-\frac{2 y^2 t'}{t^3}=1$$ Now, let $t=\frac 1{\sqrt z}$ leads to $$2yz+y^2z'=1\implies (y^2z)'=1\implies y^2z=y+C_1\implies z=\frac 1y+\frac{C_1}{y^2}$$ Back to $t$ $$t=x'=\pm \frac y {\sqrt{y+C_1}}$$ and a last integration leads to $$x=\pm\frac{2}{3} (y-2 C_1) \sqrt{C_1+y}+C_2$$
I must underline that the same solution is obtained faster using LutzL's answer.
